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Overview

(Unit Four→Heat — Second Secondary, Term 1) 🌳 Concept tree Chapter 6: Gas Laws

├── What you will learn

├── 1.1 Molecules in the three states of matter

├── 1.2 Brownian motion

├── 1.3 Intermolecular spaces

├── Lesson 2→The Variables of a Gas

├── 3.1 The experiment

├── 3.2 Boyle's law

├── 4.1 Do all gases expand equally?

├── 4.2 Charles's apparatus

├── 4.3 Charles's law

├── 5.1 The experiment

├── 5.2 Jolly's apparatus

├── 5.3 The pressure law

├── 6.1 Absolute zero from Charles's law and from the pressure law

├── 6.2 Remarks

├── 6.3 Relation between Celsius and Kelvin

├── 7.1 Combining the three laws

├── 7.2 Enrichment: the ideal gas equation

├── 7.3 Standard temperature and pressure (STP)

├── Summary and relations

Chapter 6: Gas Laws · 1 / 22

What you will learn

Brownian motion and what it tells us about gas molecules; intermolecular spaces.
The three variables of a gas (volume, pressure, temperature) and three separate experiments.
Boyle’s law (volume–pressure at constant temperature).
Charles’s law (volume–temperature at constant pressure); the coefficient of volume expansion.
The pressure law (Gay-Lussac’s law) (pressure–temperature at constant volume); Jolly’s apparatus; the coefficient of pressure increase.
Absolute zero and the Kelvin scale.
The general gas law, and the ideal gas equation PV = nRT (enrichment).
Solving problems on gases.

Chapter 6: Gas Laws · 2 / 22

1.1 Molecules in the three states of matter

The molecules of every substance are in continuous motion
but the kind of motion depends on the state:

gasrandom translational motion in all directions (and collisions);

liquidtranslational and vibrational motion;

solidvibrational motion about fixed positions only.

Chapter 6: Gas Laws · 3 / 22

1.2 Brownian motion

If we look through a microscope at the smoke rising from a burning candle, we see tiny carbon particles moving here and there in random zig-zag motion.
This motion is called Brownian motion after the Scottish botanist Robert Brown, who discovered in 1827 that pollen grains suspended in water move constantly and randomly.
Explanation.

The molecules of air (or any gas) move randomly in all directions with different speeds and collide with each other and with the smoke particles.
At any instant the number (or force) of the collisions on one side of a carbon particle may be larger than on the opposite side, so the particle is pushed a short distance in one direction.
A moment later the unbalance is in another direction.
The motion of the molecules is caused by heat and is never-ending.
Conclusion.

Gas molecules are in continuous random motion; they collide with each other and with the walls of the container.
(The collisions with the walls are the cause of gas pressure.)

Chapter 6: Gas Laws · 4 / 22

1.3 Intermolecular spaces

Experiment.

Hold a beaker full of ammonia gas (NH₃) upside down over a beaker of hydrogen chloride gas (HCl) (with a cover separating them, which is then removed).
A white cloud of ammonium chloride (NH₄Cl) forms and spreads until it fills both beakers.
Explanation.

HCl molecules, although heavier (denser), rise upward through the spaces between ammonia molecules; NH₃ molecules, although lighter, move downward through the spaces between HCl molecules.
Where they meet, they combine to form NH₄Cl, which diffuses to fill both beakers.
Conclusion.

There are relatively large spaces between gas molecules (intermolecular spaces).
This also explains why gases are easily compressed pressure pushes the molecules closer together, so the volume decreases.

Chapter 6: Gas Laws · 5 / 22

Lesson 2: The Variables of a Gas

Experiments on the expansion of a gas with heat are not simple because a gas changes its volume if the pressure, the temperature or both change.
(For solids and liquids this difficulty does not arise their compressibility is very small.)
To study a gas we have to consider three variables — volume (V), pressure (P) and temperature (T) — and do three separate experiments, in each of which the relationship between two variables is studied while the third is held constant (and the mass of the gas is fixed):

Experiment	Constant	Law
Volume against pressure	temperature	Boyle’s law
Volume against temperature	pressure	Charles’s law
Pressure against temperature	volume	Pressure law (Gay-Lussac’s law)

Chapter 6: Gas Laws · 6 / 22

3.1 The experiment

Apparatus.

A graduated glass tube (A) (like an inverted burette whose scale starts at the top) connected by a rubber tube to a second movable tube (B).
Both contain mercury.
The apparatus stands on a vertical board with levelling screws.
Tube (B) can be raised, lowered and fixed at any height.
A stopcock closes the top of tube (A).
Procedure

With the tap of (A) open, move (B) until the mercury in (A) is about halfway. Both tubes are open to the air, so the mercury is at the same level in both.
Close the tap. The trapped air has volume V₁ and pressure P₁ = P_a (atmospheric pressure in cm Hg, read from a barometer).
Raise (B) a few centimetres. The mercury level in (B) is now higher than in (A) by h. Measure the new volume V₂. The pressure is P₂ = P_a + h.
Repeat at least once more with another height of (B) above (A) (P₃, V₃).
Lower (B) so that its mercury level is lower than that in (A) by h. Measure V₄; the pressure is P₄ = P_a − h.
Repeat once more (P₅, V₅).
Plot volume (vertical axis) against 1/P (horizontal axis). The graph is a straight line through the origin.

So V ∝ 1/P.
(Keep the temperature constant wait before taking each reading so that the gas returns to room temperature, and change the pressure slowly.)

Chapter 6: Gas Laws · 7 / 22

3.2 Boyle’s law

Comparison aspect	Main idea	Keywords
Boyle’s law	At constant temperature the volume of a fixed mass of gas is inversely proportional to its pressure	constanttemperaturevolumefixed
•	V ∝ 1/P → P V = constant (fixed mass constant temperature)	constantfixedmasstemperature
or for two states	P₁V₁ = P₂V₂	P₁V₁P₂V₂
•	The graph of P against V (constant T) is a curve (a hyperbola) the graph of V against 1/P is a straight line through the origin the graph of PV against P is a horizontal line	graphagainstconstantcurve
Example 1.	A gas occupies 600 cm³ at 76 cm Hg. It is compressed at constant temperature to 400 cm³. P₂ = P₁V₁/V₂ = 76 × 600/400 = 114 cm Hg	occupiescompressedconstanttemperature

✍️ Worked examples

Comparison aspect	Main idea	Keywords
Example 2.	The pressure of a gas is tripled at constant temperature. Its volume becomes one-third	pressuretripledconstanttemperature
Example 3.	8 cm³ of air is at the bottom of a lake at a depth where the total pressure is 3 atm. The temperature is the same at the surface. At the surface (1 atm): V = 3 × 8 = 24 cm³	bottomlakedepthtotal
Example 4 (barometer-tube).	A tube holds 40 cm³ of air at atmospheric pressure of 76 cm Hg. The air is pushed to 25 cm³ by raising the movable tube. P = 76 × 40/25 = 121.6 cm Hg so h = 121.6 − 76 = 45.6 cm Hg	holdsatmosphericpressurepushed

Chapter 6: Gas Laws · 8 / 22

4.1 Do all gases expand equally?

Gases expand when heated and contract when cooled. Do equal volumes of different gases expand by different amounts?
Experiment.

Two flasks of exactly equal volume, each closed with a stopper through which passes a right-angle glass tube holding a small thread of mercury (2–3 cm long).
Fill one with oxygen and the other with air or carbon dioxide.
Place both in the same basin of water and add hot water.
The two mercury threads move equal distances.
Conclusion. Equal volumes of different gases, under the same pressure, expand by equal amounts for the same rise of temperature; so their coefficient of volume expansion is the same.
Coefficient of volume expansion of a gas at constant pressure (α_v): the increase in volume of unit volume of the gas at 0 °C when its temperature rises by one degree Celsius at constant pressure.

Chapter 6: Gas Laws · 9 / 22

4.2 Charles’s apparatus

A glass tube about 30 cm long and 1 mm wide, closed at one end, with a drop of mercury trapping a certain amount of dry air.
The tube is fixed along with a thermometer to a graduated ruler inside a cylindrical glass jacket.
Procedure

Surround the tube with melting crushed ice until the trapped air is at 0 °C. Because the tube has a uniform cross-section, the length of the air column is a measure of its volume (V₀).
Remove the ice water and pass steam through the jacket until the air reaches 100 °C. Measure the length of the air column (volume V₁₀₀).
Plot volume against temperature (°C). A straight line is obtained. If it is extended backwards, it cuts the temperature axis at −273 °C.
The coefficient of volume expansion is

αv = (V₁₀₀ − V₀) / (V₀ × 100)→For air (and for all gases) it is found to be 1/273 per °C (per kelvin).

Chapter 6: Gas Laws · 10 / 22

4.3 Charles’s law

Charles’s law (first form):

At constant pressure, the volume of a fixed mass of gas increases by 1/273 of its volume at 0 °C for every 1 °C rise in temperature.
This fraction is the same for all gases.
Thus V_t = V₀ (1 + t/273), where t is in °C.
Second form (Kelvin scale). Using similar triangles in the graph (V against t, extended to −273 °C): V₁/T₁ = V₂/T₂, i.e.
V / T = constant (fixed mass, constant pressure), so V ∝ T
At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute (Kelvin) temperature.
Example 1. A gas has volume 300 cm³ at 0 °C. At 91 °C (constant pressure): V = 300 × (273 + 91)/273 = 300 × 364/273 = 400 cm³.

Comparison aspect	Main idea	Keywords
Example 2.	2 litres of a gas at 27 °C (300 K) is heated to 127 °C (400 K) at constant pressure. V₂ = 2 × 400/300 = 2.67 litres	litresheatedconstantpressure
Example 3 (find the temperature).	A gas of volume 500 cm³ at 7 °C (280 K) is heated at constant pressure until its volume becomes 750 cm³. T₂ = 280 × 750/500 = 420 K = 147 °C	volumeheatedconstantpressure
Example 4.	Show the coefficient. A gas has 200 cm³ at 0 °C and 273.2 cm³ at 100 °C. α_v = (273.2 − 200)/(200 × 100) = 3.66 × 10⁻³ °C⁻¹ ≈ 1/273.	Showcoefficient

Chapter 6: Gas Laws · 11 / 22

5.1 The experiment

Simple experiment.

A glass flask closed with a stopper through which a U-shaped tube (A, B) containing mercury passes.
At first the mercury is at the same level (x and y) in both arms, so the gas pressure is atmospheric; temperature t₁.

Immerse the flask in warm water at t₂. The mercury falls in arm A (connected to the flask) and rises in B.
Add mercury through a funnel until the level in A returns to mark x, so the volume of the gas is again the same (constant volume).
the pressure of the gas has increased by h cm Hg when the temperature rose from t₁ to t₂.
Repeat with different gases and the same rise in temperature. Findings: (a) at constant volume, the pressure increases with temperature; (b) equal pressures of different gases increase by equal amounts for equal rises in temperature.

Coefficient of pressure increase of a gas at constant volume (β_P): the increase in the pressure of a unit pressure measured at 0 °C when the temperature rises by 1 °C at constant volume.

ΔP ∝ P₀ Δt → ΔP = β_P P₀ Δt → β_P = ΔP/(P₀ Δt) = (P₁₀₀ − P₀)/(P₀ × 100)

Chapter 6: Gas Laws · 12 / 22

5.2 Jolly’s apparatus

A spherical thin glass bulb (A) connected to a long capillary tube (B) (bent at two right angles), fixed on a vertical board that stands on a base with levelling screws.
The other end of the capillary is joined by a rubber tube to a wide tube (C) which can be moved up and down along the board; there is a graduated scale.
Procedure

Read the atmospheric pressure from a barometer.
Put mercury in the bulb’s side to take up part of the volume (about 1/7 of the bulb), so that the volume of the remaining air stays practically constant at all temperatures (mercury expands about 7 times as much as glass).
Immerse bulb (A) in a beaker of water; pour mercury into (C) until the mercury in the other branch rises to a fixed mark X.
Heat the water to boiling, wait for the temperature to become steady.
Move (C) until the mercury in the bulb-side arm returns to mark X (so the volume is again constant). The gas pressure P = P_a + h, where h is the height difference between the surfaces in the two arms.
Let the water cool to about 90 °C, adjust (C) so that the mercury returns to X, measure the temperature and height difference and calculate P again.
Repeat at several temperatures.
Plot pressure (vertical) against temperature in °C (horizontal). A straight line is obtained which, when extended backwards, cuts the temperature axis at −273 °C.
Calculate β_P = (P₁₀₀ − P₀)/(P₀ × 100). It is found to be 1/273 per °C for air and the same for all gases.

Chapter 6: Gas Laws · 13 / 22

5.3 The pressure law

Comparison aspect	Main idea	Keywords
The pressure law	At constant volume the pressure of a fixed mass of gas increases by 1/273 of its pressure at 0 °C for every 1 °C rise in temperature	constantvolumefixedmass
Second form (Kelvin scale).	From similar triangles: P₁/T₁ = P₂/T₂ i.e	similartriangles
P / T = constant	(fixed mass constant volume) so P ∝ T	fixedmassvolume
•	At constant volume the pressure of a fixed mass of gas is directly proportional to its absolute temperature	constantvolumepressurefixed
•	Note that the coefficient of volume expansion at constant pressure α_v and the coefficient of pressure increase at constant volume β_P are equal for all gases: both = 1/273 K⁻¹.	Notecoefficientvolumeexpansion
Example 1.	A gas has a pressure of 45 cm Hg at 27 °C (300 K). Its volume is constant. At 127 °C (400 K): P₂ = 45 × 400/300 = 60 cm Hg	pressurevolumeconstant

Comparison aspect	Main idea	Keywords
Example 2.	A sealed container of air is at 0 °C with pressure 80 cm Hg. It is cooled to −91 °C (182 K): P = 80 × 182/273 = 53.3 cm Hg	sealedcontainerpressurecooled
Example 3 (car tyre).	A tyre has air at 2.0 atm at 7 °C (280 K). After driving the air is at 47 °C (320 K). Assuming constant volume P = 2.0 × 320/280 = 2.29 atm. (That is why tyre pressure should be checked when the tyres are cold.)	AfterdrivingAssumingconstant
Example 4 (find the temperature).	The pressure of a gas at constant volume rises from 1 atm at 27 °C to 1.5 atm. T₂ = 300 × 1.5 = 450 K = 177 °C	pressureconstantvolumerises

Chapter 6: Gas Laws · 14 / 22

6.1 Absolute zero from Charles’s law and from the pressure law

The line on the volume–temperature (°C) graph, when extended, cuts the temperature axis at −273 °C (the volume appears to vanish there).
The line on the pressure–temperature (°C) graph (Jolly’s experiment), when extended, also cuts the temperature axis at −273 °C (the pressure appears to vanish).
Absolute zero (zero kelvin) is therefore the lowest theoretically attainable temperature, −273 °C; the temperature at which the volume of an ideal gas becomes zero at constant pressure, or the pressure of an ideal gas becomes zero at constant volume.
Example of the method. For a gas whose volume is 273 cm³ at 0 °C and 373 cm³ at 100 °C, the line V = 273 + t reaches V = 0 at t = −273 °C.

Chapter 6: Gas Laws · 15 / 22

6.2 Remarks

Redrawn against the absolute temperature T (kelvin), the V–T and P–T graphs are straight lines through the origin: V = 0 and P = 0 at 0 K.
In practice, a real gas does not reach these values on strong cooling it liquefies (and may solidify) and no longer obeys the gas laws. An ideal gas is defined as a gas whose volume (at constant pressure) or pressure (at constant volume) would become zero at absolute zero, so that it obeys the gas laws at all temperatures.
The temperature on the Kelvin scale is always positive; Celsius temperature can be positive or negative.

Chapter 6: Gas Laws · 16 / 22

6.3 Relation between Celsius and Kelvin

−273 °C ↔ 0 K
0 °C ↔ 273 K
100 °C ↔ 373 K. Hence

T (K) = t (°C) + 273

Examples human body 37 °C = 310 K; room 27 °C = 300 K; 100 K = −173 °C. A change of 1 °C equals a change of 1 K.
The more exact conversion uses 273.15: T (K) = t (°C) + 273.15, so absolute zero is −273.15 °C.

Chapter 6: Gas Laws · 17 / 22

7.1 Combining the three laws

For a fixed mass of gas:

Comparison aspect	Main idea	Keywords
Boyle	V ∝ 1/P (T constant)	constant
Charles	V ∝ T (P constant) Combining: V ∝ T/P → PV/T = constant	constantCombining
•	For two states of the same mass of gas	statessamemass
P₁V₁/T₁ = P₂V₂/T₂	(T in kelvin P in any consistent unit V in any consistent unit)	kelvinconsistentunit
Special cases	T constant → Boyle’s law P constant → Charles’s law V constant → pressure law	constantBoyleCharlespressure

Chapter 6: Gas Laws · 18 / 22

7.2 Enrichment: the ideal gas equation

In general, PV = nRT, where

n = number of moles of the gas,

P in pascals, V in m³, T in kelvin,

Comparison aspect	Main idea	Keywords
R = 8.31 J/(mol·K)	the universal gas constant. A mole is the amount of a substance whose mass in grams equals its molecular mass: one mole of oxygen (O₂) has a mass of 32 g nitrogen (N₂) 28 g carbon dioxide (CO₂) 44 g	universalconstantmoleamount
Example	the volume of one mole of any gas at STP is V = nRT/P = 1 × 8.31 × 273/1.013 × 10⁵ = 0.0224 m³ = 22.4 litres	volumemole0224litres

Chapter 6: Gas Laws · 19 / 22

7.3 Standard temperature and pressure (STP)

Comparison aspect	Main idea	Keywords
STP	0 °C (273 K) and 76 cm Hg (1 atm = 1.013 × 10⁵ Pa)
Example 1 (to STP).	A gas has volume 450 cm³ at 27 °C and 60 cm Hg. Find its volume at STP	volumeFind
P₁V₁/T₁ = P₂V₂/T₂	60 × 450/300 = 76 × V₂/273 → V₂ = 60 × 450 × 273/(300 × 76) = 323 cm³
Example 2 (find T).	A gas of 3 L at 2 atm and 27 °C changes to 6 L at 1.5 atm. T₂ = T₁ P₂V₂/(P₁V₁) = 300 × (1.5 × 6)/(2 × 3) = 300 × 1.5 = 450 K = 177 °C	changesP₂V₂P₁V₁
Example 3 (find P).	A fixed mass of gas at 76 cm Hg 300 K and 500 cm³ is compressed to 250 cm³ and heated to 360 K: P₂ = 76 × 500 × 360/(300 × 250) = 182.4 cm Hg	fixedmasscompressedheated
Example 4 (gas that escapes).	An open flask of air is heated from 7 °C (280 K) to 147 °C (420 K) at constant pressure. The volume of the flask is constant so the amount of gas ∝ 1/T. Fraction left = 280/420 = 2/3 fraction that escaped = 1/3 = 33.3 % of the original amount	openflaskheatedconstant

✍️ Worked examples

Example 5 (mixture of two gases — partial pressures).
Gas A 6 L at 20 cm Hg.
Gas B 4 L at 30 cm Hg.
Both are transferred (at constant temperature) to a 3-L container.
A: P_A V_A = P’_A × 3 → P’_A = 20 × 6/3 = 40 cm Hg.
B: P’_B = 30 × 4/3 = 40 cm Hg.
The pressure of a mixture equals the sum of the pressures each gas would produce alone: P = 40 + 40 = 80 cm Hg.
Example 6 (ideal gas equation). Two moles of a gas at 300 K in a 0.01 m³ container: P = nRT/V = 2 × 8.31 × 300/0.01 = 4.99 × 10⁵ Pa (about 4.9 atm).
Example 7 (helium balloon). A helium balloon rises in the air. As it rises the surrounding pressure falls (and the temperature falls less than the pressure), so by PV/T = constant its volume increases, until it may burst.

Chapter 6: Gas Laws · 20 / 22

Summary and relations

Gas molecules are in continuous random motion, colliding with each other and with the walls; there are spaces between molecules (compressibility). Evidence: Brownian motion, diffusion of NH₃ and HCl.
Boyle PV = const (T const).
Charles V/T = const (P const); V increases by 1/273 of V₀ per °C; αv = (V₁₀₀ − V₀)/(V₀ × 100) = 1/273 K⁻¹.
Pressure law P/T = const (V const); P increases by 1/273 of P₀ per °C; βP = (P₁₀₀ − P₀)/(P₀ × 100) = 1/273 K⁻¹.
T(K) = t(°C) + 273; absolute zero = −273 °C.
General gas law P₁V₁/T₁ = P₂V₂/T₂; PV = nRT with R = 8.31 J/(mol·K).
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Test yourself on this lesson

The lesson’s question bank is split across the points above; each point has its own questions underneath.
To work through the whole bank on one page:

Chapter 6: Gas Laws · 21 / 22

Part 22

📖

📌 Comprehensive Scientific Terminology & Definitions:

Comparison aspect	Main idea	Keywords
Boyle’s Law	At constant temperature the volume of a fixed mass of an ideal gas is inversely proportional to its pressure: P ∝ 1 / V meaning P₁ V₁ = P₂ V₂ = constant	constanttemperaturevolumefixed
Charles’s Law	At constant pressure the volume of a fixed mass of an ideal gas is directly proportional to its absolute temperature in Kelvin: V ∝ T meaning V₁ / T₁ = V₂ / T₂ = constant	constantpressurevolumefixed
Pressure Law (Gay-Lussac’s Law)	At constant volume the pressure of a fixed mass of an ideal gas is directly proportional to its absolute temperature in Kelvin: P ∝ T meaning P₁ / T₁ = P₂ / T₂ = constant	constantvolumefixedmass
Absolute Zero (0 Kelvin / -273.15°C)	The theoretical temperature at the volume (at constant P) or pressure (at constant V) of an ideal gas would extrapolate to zero molecular kinetic energy reaches its minimum possible value	theoreticaltemperaturevolumeconstant
Ideal Gas Equation	Equation of state combining all three empirical gas laws: P V = n R T where n is the number of moles and R is the universal gas constant (8.314 J/mol·K)	statecombiningthreeempirical

⚖️ Governing Scientific Laws, Principles & Mathematical Formulas:

Comparison aspect	Main idea	Keywords
Boyle’s Law (T = const)	P₁ V₁ = P₂ V₂
Charles’s Law (P = const)	V₁ / T₁ = V₂ / T₂ (where T = t°C + 273.15)
Pressure Law (V = const)	P₁ / T₁ = P₂ / T₂
General Combined Gas Law	(P₁ V₁) / T₁ = (P₂ V₂) / T₂
Ideal Gas Law	P V = n R T = (m / M) R T
Volume Expansion Coefficient	α_v = (Vt – V0) / (V0 × t) = 1 / 273.15 K⁻¹

🔍 In-Depth Explanations, Deductions & Core Concepts:

Comparison aspect	Main idea	Keywords
•	Why do all gases have the identical thermal expansion coefficient (α_v = 1/273)? Because intermolecular spacing in gases is enormous and cohesive intermolecular forces are negligible so all ideal gases expand identically regardless of chemical identity	gasesidenticalthermalexpansion
Why can gas volume never actually reach zero at absolute zero?	Because real gases condense into liquids and freeze into solids long before reaching 0 K as attractive van der Waals forces overcome kinetic energy	Becauserealgasescondense
Why do car tires sometimes burst on hot highways?	Friction raises the temperature of the enclosed air. Since tire volume is approximately rigid (constant V) air pressure climbs proportionally with absolute temperature (P ∝ T) until structural failure occurs	Frictionraisestemperatureenclosed

Gas Law	Constant Parameter	Proportionality	Governing Formula
Boyle’s Law	Temperature (T)	Inverse: P ∝ 1/V	P₁ V₁ = P₂ V₂
Charles’s Law	Pressure (P)	Direct: V ∝ T(K)	V₁ / T₁ = V₂ / T₂
Pressure Law	Volume (V)	Direct: P ∝ T(K)	P₁ / T₁ = P₂ / T₂

💡 High-Yield Exam Traps & Key Takeaways:

Always convert Celsius to KelvinT(K) = t(°C) + 273.15.

For gas mixtures at constant TPtotal Vtotal = P₁ V₁ + P₂ V₂.

Chapter 6: Gas Laws · 22 / 22

Curriculum

Physics — Second Secondary

Physics — Second Secondary

Physics — Second Secondary 2026/2027

Chapter 6: Gas Laws

├── What you will learn

├── 1.1 Molecules in the three states of matter

├── 1.2 Brownian motion

├── 1.3 Intermolecular spaces

├── 3.1 The experiment

├── 3.2 Boyle's law

├── 4.2 Charles's apparatus

├── 4.3 Charles's law

├── 5.1 The experiment

├── 5.2 Jolly's apparatus

├── 5.3 The pressure law

├── 6.2 Remarks

├── 6.3 Relation between Celsius and Kelvin

├── 7.1 Combining the three laws

├── 7.2 Enrichment: the ideal gas equation

├── 7.3 Standard temperature and pressure (STP)

├── Summary and relations

✍️ Worked examples

T (K) = t (°C) + 273

n = number of moles of the gas,

P in pascals, V in m3, T in kelvin,

✍️ Worked examples

Test yourself on this lesson

📌 Comprehensive Scientific Terminology & Definitions:

⚖️ Governing Scientific Laws, Principles & Mathematical Formulas:

🔍 In-Depth Explanations, Deductions & Core Concepts:

Well done! You finished this lesson

P in pascals, V in m³, T in kelvin,